STATEMENT: So let us assume $V$ is finite dimensional vector space with an inner product $\langle.,.\rangle_V$. We map $V$ isomorphically into its dual space $V^*$ by sending $v\mapsto \langle v,.\rangle$. We call this map $\phi$. Furthermore, if $a,b\in V^*$ we define an inner product on $V^*$ by $\langle a,b\rangle_{V^*}=\langle \phi^{-1}(a),\phi^{-1}(b)\rangle_V$.
Now suppose $\left\{e_i\right\}$ is a basis for $V$ and $\left\{e^i\right\}$ is the dual basis. Since our map sends $\phi(e_i)=\sum_{j=1}^nb_{ij}e^j$ where $b_{ij}=\langle e_i,e_j\rangle_V$ we want to show that if $b^{ij}=\langle e^i,e^j\rangle_{V^*}$ then the matrix $(b^{ij})$ is the inverse matrix of $(b_{ij})$.
QUESTION: I can't seem to show that $\langle e^i,e^j\rangle=\frac{1}{|(b_{ij})|}C_{ij}$. Showing this would be sufficient to proving the statement. I have posted a partial sketch of my proof below, I am wondering if anyone has suggestions or hints on what I should do next.
PROOF:Let us assume an orthonormal basis $\left\{e_i'\right\}$ on V(such a basis exists by Gram-Schmidt process). Thus it follows that $b_{ij}=1$ iff $i=j$ and 0 otherwise. Since the inner product passes onto the image $f\left\{e_i'\right\}$, it follows that the image is orthonormal so $(b^{ij})=Id$. Thus $(b^{ij})=(b_{ij})^{-1}$trivially. Now we use basis transformation matrices to realize for an arbitrary basis $\left\{e_i\right\}$ on $V$. We can take $B$ to be the basis transformation matrix from $\left\{e_{ij}'\right\}$ to $\left\{e_{ij}\right\}$. Thus, $(b_{ij})= B B^T$. It is sufficient for us to show that $(b^{ij})=(B^T)^{-1}B^{-1}$.
I won't comment on your approach, which seems rather roundabout; to begin with I don't understand which $g_{i,j}$ and $C_{i,j}$ you are talking about.
You just need to compute $b^{i,j}\overset{\mathrm{def}}=\langle e^i,e^j\rangle_{V^*}$. Since the purpose is to show this is related to the inverse matrix $B^{-1}$ of $B=(b_{i,j})_{i,j=1}^n$ start with interpreting $B$ as the matrix of a linear map. Since your inner product is symmetric, so is the matrix $B$. Then you may as well write $\phi(e_j)=\sum_{i=1}^nb_{i,j}e^i$ instead of what is in the question; this says that the columns of $B$ express the images $\phi(e_j)$ in the chosen dual basis of$~V^*$, so $B$ is the matrix of $\phi$ with respect to the chosen bases of $V$ and of $V^*$. But then $B^{-1}$ is the matrix of $\phi^{-1}$ with respect to the chosen bases of $V^*$ and of $V$. Putting $B^{-1}=(c_{i,j})_{i,j=1}^n$ one then has $$ \phi^{-1}(e^j)=\sum_{i=1}^nc_{i,j}e_i \qquad\text{for $j=1,\ldots,n$.} $$
Now we come to actually computing $\langle e^i,e^j\rangle_{V^*}$. You defined $\langle a,b\rangle_{V^*}=\langle \phi^{-1}(a),\phi^{-1}(b)\rangle_V$, but you can do a bit better than apply $\phi^{-1}$ on both sides. Recall that by definition of$~\phi$ one has for all $v,w\in V$ that $\langle v, w\rangle=\phi(v)(w)$, so applying that to $v=\phi^{-1}(a)$ one gets $\langle a,b\rangle_{V^*}=a(\phi^{-1}(b))$ (that is apply the linear form $a\in V^*$ to $\phi^{-1}(b)\in V$). Now things are easy: $$ \langle e^i,e^j\rangle_{V^*}=e^i(\phi^{-1}(e^j)) =e^i\Bigl(\sum_{k=1}^nc_{k,j}e_k\Bigr) =\sum_{k=1}^nc_{k,j}e^i(e_k) =\sum_{k=1}^nc_{k,j}\delta_{i,k} =c_{i,j}, $$ by definition of dual bases, as desired.